math/check+ help! conic in standard form

• Dec 5th 2006, 12:41 PM
george93
math/check+ help! conic in standard form
Hi guys!

use the graph 4x^2-y^2-8x-4y+16=0 to do the following:
a. write the equation in standard form

I did but could you double check my work?

4x^2-y^2-8x-4y+16=0
4x^2-y^2-8x-4y=-16
4(x^2-2x) -(Y^2+4y) =-16
4[(X^2-2x+1)-1] - [(Y^2+4y+4)-4] =-16
4[(x-1)^2-1] - [(Y+2)^2-4] =-16
4(x-1)^2-4 - (y+2)^2+4 =-16
4(x-1)^2- (y+2)^2= -16

4(x-1)^2- (y+2)^2= -16
-16

(x-1)^2 - (y+2)^2 = 1
-4 ....... 16

I think I did that right! but not sure!

there is another part to it:
locate the center and the vertices of this curve

The center is:
(-2,1)

how would I find out the vertices of the curve?

• Dec 5th 2006, 12:44 PM
ThePerfectHacker
Quote:

Originally Posted by george93
H
4(x-1)^2- (y+2)^2= -16
-16

(x-1)^2 - (y+2)^2 = 1

-4 rtetetewt erwerwqerwerwq 16
!

It is correct but ugly like the Uzi.
You have,
$\displaystyle 4(x-1)^2-(y+2)^2=-16$
Multiply by (-1),
$\displaystyle (y+2)^2-4(x-1)^2=16$
Thus,
$\displaystyle \frac{(y+2)^2}{4^2}-\frac{(x-1)^2}{2^2}=1$
This is a hyperbola.